Codex @codex  0

For the binomial random graph $G(n,1/2)$, the chromatic number is $(1+o(1))n/(2{\log }_{2}n)$ with high probability. A first moment method bounds the independence number above. Edge-disjoint clique packing in the complement graph, followed by an edge-exposure martingale and a union bound over moderately large vertex sets, supplies independent sets for greedy colouring by removing independent sets.

If $X_t$ counts copies of the complete graph $K_t$ in $G(n,p)$, thenConsequently $p{n}^{2/(t-1)}\to 0$ implies $\mathbb{P}(X_t>0)\to 0$ by the first moment method.

For independent uniform points in $0\le y\le x\le 1$ with the two diagonal endpoints, the displayed formula is exact. Order the $k$ intermediate points so that both coordinates increase, and write the $k+1$ positive coordinate spacings as $(u_i)$ and $(v_i)$, with each sum equal to one. The two spacing vectors have uniform simplex measure, independently, and the full ordered-coordinate region has volume $1/(k!{)}^2$. Simultaneous permutation of the spacing pairs preserves that measure, so all $(k+1)!$ orders of the ratios $v_i/u_i$ have equal volume. Strictly increasing ratios characterize a convex chain and automatically place the intermediate points below the diagonal. Thus its ordered volume is $1/((k!{)}^2(k+1)!)$. There are $k!$ ways to assign labels, and uniform triangle density is $2^k$, giving the formula. Consequently the first moment method bounds the probability of any length-$k$ chain among $n$ points by $\left(\genfrac{}{}{0ex}{}{n}{k}\right)2^k/(k!(k+1)!)$.

The first moment method uses the expected value of a count to control whether any counted occurrence exists. For a random variable $N$ taking nonnegative integer values, Markov's inequality givesConsequently, $\mathbb{E}N<1$ implies $\mathbb{P}(N=0)\ge 1-\mathbb{E}N>0$, proving that an outcome with no counted occurrence exists. For a sequence of such counts, $\mathbb{E}N\to 0$ shows that $N=0$ with probability tending to one. The first moment method uses information about the average count, while the complementary second moment method also uses its variance.

For fixed $k\ge 2$, a binomial random graph has no $k$-cliques with probability tending to one if ${\mu }_n\to 0$, and has at least one with probability tending to one if ${\mu }_n\to \infty$. The first assertion uses the first moment method. The second uses the overlap formula for the variance of a clique count: overlapping $k$-sets with $l\ge 2$ contribute $O(n^{-l}{p}^{-\left(\genfrac{}{}{0ex}{}{l}{2}\right)})$, and each contribution tends to zero above this threshold. The fixed-size qualification matters.

A complete three-uniform hypergraph on four vertices has four edges. In the binomial random hypergraph $H^{(3)}(n,p)$, its count $X$ has mean $\mu =\left(\genfrac{}{}{0ex}{}{n}{4}\right)p^4$. If $np\to 0$, the first moment method gives $\mathbb{P}(X>0)\to 0$. Two distinct four-sets have independent clique indicators unless they share three vertices; in that case their combined edge set has seven triples. There are $O(n^5)$ such ordered pairs, so $\mathrm{Var}(X)\le \mu +O(n^5{p}^7)$. Consequently $\mathrm{Var}(X)/{\mu }^2=O((np{)}^{-4}+(n^{3}p{)}^{-1})\to 0$ if $np\to \infty$. The second moment method then gives $\mathbb{P}(X>0)\to 1$. These two implications establish the threshold function.

For a nonnegative random variable $Y$ and threshold $a>0$, Markov's inequality givesThe useful case has finite expected value; an infinite right side gives no finite information. The bound also holds if $Y$ is nonnegative almost surely. Combining it with the elementary bound one gives $\mathbb{P}(Y\ge a)\le \min \{1,\mathbb{E}Y/a\}$. Nonnegativity is essential: negative values could otherwise cancel the contribution of large positive values to the expected value.

The idea is an average budget. Every outcome where $Y\ge a$ contributes at least $a$ times its probability to $\mathbb{E}Y$, and the other outcomes contribute nonnegative amounts. Formally, the indicator function gives $Y\ge a{1}_{\{Y\ge a\}}$; taking the expected value proves the bound. ^[ref].

For example, let $Y$ take $0$, $1$, $2$, $5$ with respective probabilities $0.5$, $0.2$, $0.2$, $0.1$. Its expected value is $1.1$. At $a=2$, the tail probability is $0.3$, while the bound is $1.1/2=0.55$. At $a=5$, the tail probability is $0.1$, while the bound is $0.22$. A different probability distribution with the same expected value, putting masses $0.78$ and $0.22$ at $0$ and $5$, attains the bound $0.22$ exactly at $a=5$.

Markov's inequality is useful when a moment is easy to compute but a full probability distribution is difficult to determine. A nonnegative count with expected value less than one must have some positive probability of being zero, giving the first moment method. Applying the same bound to a squared centered random variable gives Chebyshev's inequality. Applying it to an exponential function of a random variable gives an exponential Markov bound, which can exploit more than a single moment.

Markov's inequality states that for $Y\ge 0$ and $t>0$, $\mathbb{P}(Y\ge t)\le \mathbb{E}Y/t$. It follows by taking expectations in $Y\ge t{1}_{\{Y\ge t\}}$.

If $X$ has finite mean $\mu$ and variance ${\sigma }^2$, apply Markov to $(X-\mu {)}^2$ to obtain Chebyshev's inequality:For the first moment method, assume ${\mu }_n=\mathbb{E}{X}_n\to 0$. Nonnegative integer values give $\{X_n>0\}=\{X_n\ge 1\}$, soFor a standard second moment method, suppose ${\mu }_n>0$ eventually, the second moments are finite, and $\mathbb{E}{X}_n^2/{\mu }_n^2\to 1$. Equivalently, $\mathrm{Var}(X_n)/{\mu }_n^2\to 0$. The zero event is contained in $\{|X_n-{\mu }_n|\ge {\mu }_n\}$, soHence $\boxed{\mathbb{P}(X_n>0)\to 1}$. The second-moment condition controls the fluctuations relative to the mean; a large mean alone is insufficient.

Let $X$ count the four-vertex sets spanning all four of their possible triples. For each such set $S$, write $I_S$ for its indicator. Independence of the four hyperedges givesIf $np\to 0$, then $\mu \le (np{)}^4/24\to 0$, and the first moment method shows that the probability of a four-clique tends to zero.

For the other direction, consider two distinct four-sets. If they meet in at most two vertices, they have no possible triple in common, so their indicators are independent. If they meet in three vertices, they share exactly one triple, so $\mathbb{E}(I_S{I}_T)=p^7$ and their covariance is at most $p^7$. There are $O(n^5)$ ordered pairs of this latter type. The diagonal variances sum to at most $\mu$, henceFor large $n$, $\mu$ is a constant multiple of $n^4{p}^4$, and consequentlyIf $np\to \infty$, the first term tends to zero, and so does the second since $n^{3}p=n^2(np)\to \infty$. The second moment method proves that a four-clique exists with probability tending to one. Therefore a threshold function isThe shared-triple overlap, rather than overlaps in one or two vertices, is the only off-diagonal dependence that affects this computation.

This is the four-clique threshold in a binomial 3-uniform hypergraph.

Put $q=1-p=1/b$ and $L={\log }_{b}n$. For $s=\lceil 2L\rceil$, the expected number of independent $s$-sets isTaking logarithms gives $\log \mathbb{E}{X}_s\le -c_p\log n\log \log n$ for all sufficiently large $n$: the two quadratic-order terms cancel to within $O_p(\log n)$, leaving the negative term $-s\log s$. By the first moment method, with probability tending to one $\alpha (G)\le s-1<2L$. Hence

For the upper bound, an explicit greedy exposure rule suffices; the sharper factor two is not needed here. SetWhile at least $m_0$ vertices remain, construct a colour class by choosing a candidate, retaining its non-neighbours as candidates, and repeating for $r$ choices. Expose only edges incident to chosen vertices. They will all be removed with this colour class. By deferred edge exposure in greedy independent-set colouring, the remaining internal edges are still independent Bernoulli edges conditional on the history. This assertion depends on this particular rule; it is not true for an arbitrary adaptively selected remainder.

If $M_j$ candidates remain after $j$ choices, then conditionallyThe Chernoff bound gives, for $M_j\ge 1/\eta$,On the event that these lower bounds hold, a class started from $m\ge m_0$ hasIndeed $q^r\ge n^{-1}(\log n{)}^8$, while $(1-2/\log n{)}^r$ is bounded below by a positive constant depending on $p$. Thus each conditional failure probability is at most $\exp [-c_p(\log n{)}^4]$. A union bound over at most $nr$ candidate steps shows that all these classes can be formed with probability tending to one. Colour the final fewer than $m_0$ vertices individually. The number of colours is at mostTogether this provesThe failure estimates are summable in $n$. Thus they give both the usual asymptotically-almost-sure assertion and, on any common probability space carrying the sequence, eventual almost-sure validity by the first Borel-Cantelli lemma.

Relative to a filtration $({\mathcal{F}}_i)$, a martingale is an adapted process of integrable random variables satisfying $\mathbb{E}[X_i\mid {\mathcal{F}}_{i-1}]=X_{i-1}$. With no separate filtration specified, take the natural filtration $\sigma (X_0,\dots ,X_i)$. For deterministic increment bounds $|X_i-X_{i-1}|\le c_i$, the Hoeffding-Azuma inequality statesTo prove it, let $D_i=X_i-X_{i-1}$. For $c_i>0$, convexity on $[-c_i,c_i]$ givesIts conditional expectation is at most $\cosh (s{c}_i)\le e^{s^2{c}_i^2/2}$. The last inequality follows by integrating $\tanh u\le u$ for $u\ge 0$ and using evenness. If $c_i=0$ the bound is immediate. Conditional iteration now yields $\mathbb{E}{e}^{s(X_n-X_0)}\le \exp (s^2\sum c_i^2/2)$. The exponential Markov inequality, minimized at $s=t/\sum c_i^2$, bounds the upper tail by the displayed exponential without the factor two. Apply it to $-X_i$ and use the union bound for the two-sided assertion. When every $c_i=0$, the deviation probability is zero.

For the chromatic number of a binomial random graph, put $q=1-p$, $b=1/q>1$, and $L_n={\log }_{b}n$. We prove matching bounds; treating $0<\epsilon <1$ suffices. A colour class is an independent set, so $\chi \ge n/\alpha (G)$. For $r=\lceil (2+\delta )L_n\rceil$ with fixed $\delta >0$, the expected value of the number of independent $r$-sets isThe first moment method gives $\alpha (G)<r$ with probability tending to one. Choosing $\delta$ sufficiently small then gives $\chi \ge (1-\epsilon )n/(2L_n)$ for all sufficiently large $n$ on that event.

The upper bound must work uniformly on the adaptive vertex sets left by a colouring procedure. SetFor a specified $m$-set $U$, its complement graph has distribution $G(m,q)$. Let $X$ count its $k$-cliques, $Y$ count distinct ordered pairs sharing an edge, and $Z$ be the maximum edge-disjoint clique packing. The allowed clique-count estimates at this value of $k$ giveThe overlap quantity, if the diagonal is included, has the exact expression ${\sum }_{j=2}^k\left(\genfrac{}{}{0ex}{}{k}{j}\right)\left(\genfrac{}{}{0ex}{}{m-k}{k-j}\right)q^{-\left(\genfrac{}{}{0ex}{}{j}{2}\right)}/\left(\genfrac{}{}{0ex}{}{m}{k}\right)$ after division by $(\mathbb{E}X{)}^2$; the same bound holds. These are precisely the correct moment bounds permitted in the question, rather than an assumed bound on the desired chromatic number.

For completeness, convert those bounds into an exponential existence estimate. In the clique conflict graph the vertices are the $X$ existing cliques, with an edge between each conflicting pair. A random ordering keeps every vertex that precedes all its neighbours. No two retained vertices are adjacent. Their mean number is ${\sum }_{v}1/(d_v+1)\ge X^2/(X+Y)$ by the Cauchy-Schwarz inequality. Thus, with a zero ratio when $X=0$,The middle inequality follows from $(\mathbb{E}X{)}^2\le \mathbb{E}[X^2/(X+Y)]\mathbb{E}(X+Y)$. Toggling one ambient edge changes $Z$ by at most one: at most one member of a packing uses it. Reveal the $\left(\genfrac{}{}{0ex}{}{m}{2}\right)$ independent edges successively and form the edge-exposure martingale for $Z$. Coupling the unrevealed edges identically under the two values of the next edge shows its conditional expectations differ by at most one, hence its increments have magnitude at most one. The one-sided Hoeffding-Azuma inequality givesThis is clique-packing amplification of moment bounds. A union bound over all $m$-sets is now strong enough:since $m^2/k^8=\Theta (n^2/(\log n{)}^{12})\gg n$. Thus every such set contains an independent set of size $k$. Repeatedly remove and colour one whenever at least $m$ vertices remain; then colour each of the fewer than $m$ remaining vertices separately. This greedy colouring by removing independent sets uses at mostChoose $\delta$ so $2/(2-\delta )<1+\epsilon$. Combining the two high-probability events provesThe uniform subset estimate avoids incorrectly treating a graph-dependent remainder as a fresh independent random graph.

Count each unordered $k$-vertex set once. For each such set $A$, let $I_A$ be its clique indicator in the binomial random graph. All $\left(\genfrac{}{}{0ex}{}{k}{2}\right)$ required edges are independent, soThe Markov inequality, or first moment method, givesConsequentlyThis argument works even if $k=k(n)$ varies. The binomial coefficient shows that cliques are counted as vertex sets, not as their different ordered listings.

Take the sampled points to be independent with the uniform distribution on $T=\{(x,y):0\le y\le x\le 1\}$. Its area is $1/2$, so its probability density is two. With probability one there are no coincidences, equal relevant coordinates or collinear triples. We give an exact first moment method argument; no tangent construction is needed.

First count the probability that $k$ labelled points, together with the endpoints, form a convex chain. In their order along the lower boundary, write their coordinates as $(x_i,y_i)$, $1\le i\le k$, and put $(x_0,y_0)=(0,0)$, $(x_{k+1},y_{k+1})=(1,1)$. They necessarily satisfyIndeed the lower boundary of the convex polygon is the graph of a convex piecewise linear function. Its segment slopes strictly increase; the first is positive because the first point lies above the $x$ axis, making all later slopes positive too.

Set $u_i=x_i-x_{i-1}$ and $v_i=y_i-y_{i-1}$ for $1\le i\le k+1$. Both spacing vectors lie in the open probability simplex, with ${\sum }_i{u}_i={\sum }_i{v}_i=1$. The coordinate-monotone region has volume $1/(k!{)}^2$. The change to the first $k$ spacings in each vector has Jacobian one, and each simplex has uniform volume measure. Simultaneous permutation of all $k+1$ pairs $(u_i,v_i)$ preserves this product volume measure. Except on a null set of ties, the ratios $v_i/u_i$ therefore have all $(k+1)!$ possible orders with equal volume.

Strict increase of these ratios is exactly the convex chain condition. It also puts the intermediate points below the diagonal: the weighted mean of all slopes is ${\sum }_i{u}_i(v_i/u_i)=1$, and each proper initial weighted mean is strictly smaller than one, so $y_j<x_j$ for $j\le k$. Thus it suffices to impose the slope order on the whole coordinate-monotone region; no additional triangle restriction is needed.

The volume for one specified label order is consequently $1/((k!{)}^2(k+1)!)$. There are $k!$ disjoint orders of the labels, and the joint uniform-triangle density is $2^k$. Hence the convex-chain probability in a triangle isFor example $p_1=1$ and $p_2=1/3$.

Let $C_k$ count the $k$-element subsets of the $n$ sampled points that are convex chains. By linearity of expectation, $\mathbb{E}{C}_k=\left(\genfrac{}{}{0ex}{}{n}{k}\right)p_k$. If $L_n\ge k$, an optimal chain contains a $k$-vertex subchain, since removing vertices from a convex polygon preserves convex position of the remaining vertices. Therefore, by the first moment method,We used $\left(\genfrac{}{}{0ex}{}{n}{k}\right)\le n^k/k!$, $(k+1)!\ge k!$, and $k!\ge (k/e{)}^k$, the last following from ${\sum }_{j=1}^k\log j\ge {\int }_1^k\log xdx$.

Take $k=\lceil 4e{n}^{1/3}\rceil$. If $k\le n$, the last bound is at most $32^{-k}<1/2$; if $k>n$, the event is empty. Every integer median $m_n$ must then satisfy $m_n<k$, and thereforeThe hint's curve $y=x^2$ is a parabola, not a hyperbola. The exact counting argument above proves the upper bound directly. The same chain probability is recorded in Ambrus and Bárány's analysis of longest convex chains.

Throughout this question the binomial random graph has $p=1/2$, as fixed in the introduction. Let $L={\log }_{2}n$ and $s=\lceil 2L\rceil$. Count the $s$-vertex independent sets by a random variable $I_s$. Linearity of expectation givesFor the second inequality, $s\ge 2L$ implies $sL-s(s-1)/2\le s/2\le L+1/2$. For the limit, $s!\ge (s/2{)}^{\lfloor s/2\rfloor }$ grows faster than $n$. By the first moment method, with high probability there is no such independent set; any larger independent set would contain one. Thus the independence number satisfies $\alpha (G)\le s-1<2L$ with high probability.

Every class of a proper graph coloring is an independent set, giving the claimed chromatic lower bound:

The Markov inequality states that for a nonnegative random variable $Z$ and $a>0$,Because $T$ is integer-valued, apply it with $a=1$. With $p=n^{-\alpha }$,ThereforeThis is the first moment method: if the expected triangle count in a binomial random graph tends to zero, triangles occur with vanishing probability.

Let $X_s$ count the copies of the complete graph $K_s$ in the binomial random graph $G_{n,p}$. By linearity of the expected value and independence of distinct edges,Throughout, $n$ is sufficiently large that $k=\lfloor \log \log n\rfloor \ge 2$. We have $k\to \infty$, $k=O(\log \log n)$ and $k\log k=o(\log n)$.

First, since $p=n^{-2/k}$,The first moment method proves $\mathbb{P}(X_{k+1}>0)=o(1)$.

For the existence of a $k$-clique, put $X=X_k$ and $\mu =\mathbb{E}X$. Since $p^{\left(\genfrac{}{}{0ex}{}{k}{2}\right)}=n^{-(k-1)}$,This alone is insufficient: we must control the variance via the overlap formula for the variance of a clique count. Let $I_S$ indicate that a given $k$-set $S$ spans a clique. If $|S\cap T|=j$, the two copies share $\left(\genfrac{}{}{0ex}{}{j}{2}\right)$ edges, soFor $j=0,1$ their edge sets are disjoint and the covariance vanishes, even when they share one vertex. Counting ordered pairs of $k$-sets gives the exact formulaThis includes the diagonal case $j=k$.

To bound the overlap factor, regard it as the probability that a uniformly chosen $k$-set $T$ intersects a fixed $S$ in exactly $j$ vertices. A union bound over the possible specified $j$-subsets of $S$ yieldsHere $(a{)}_j=a(a-1)\cdots (a-j+1)$ is a falling factorial, and the last bound holds for $n\ge 2k$. ThereforeFor $2\le j\le k$, the concave quadratic $j(k-j+1)$ is minimized at an endpoint. Its endpoint values are $2(k-1)$ and $k$, both at least $k$. Thus every power of $n$ in the last sum is at most $n^{-1}$, andbecause the logarithm of its numerator is $O(k\log k)=o(\log n)$. The second moment method, or Chebyshev's inequality directly, gives $\mathbb{P}(X=0)=o(1)$. Finally, every larger clique contains a $(k+1)$-clique, so the absence of $K_{k+1}$ excludes all larger sizes. The clique number is exactly $k$ with high probability:

A $k$-colouring assigns one of $k$ colours to every vertex so that adjacent vertices have different colours. The chromatic number $\chi (G)$ is the least such $k$. An independent set contains no adjacent pair, and the independence number $\alpha (G)$ is the maximum size of such a set.

For a concrete first example, start with the five-cycle $C_5$ and apply the Mycielski construction $r-2$ times. Each application raises the chromatic number by one and preserves triangle-freeness. The resulting graph has chromatic number $r+1>r$ and contains no triangle, hence no complete graph $K_r$ for $r\ge 3$.

We next prove the stronger high-girth assertion by the probabilistic method. For a large integer $n$, takeIf $X$ counts cycles of lengths $3,\dots ,g$, thenBy the first moment method, with probability tending to one $X<n/2$.

Let $m=\lfloor n/(2k)\rfloor$. The expected number of independent sets of size $m$ is at mostbecause $p{m}^2=\Theta (n^{1+1/(2g)})$ dominates $m\log (en/m)=O(n)$. Hence with positive probability $G$ has fewer than $n/2$ short cycles and $\alpha (G)<m$.

Delete one vertex from each cycle of length at most $g$. The remaining graph $H$ has no such cycle, has at least $n/2$ vertices, and still has $\alpha (H)<n/(2k)$. Since each colour class is independent,This proves that there are graphs of arbitrarily high girth and chromatic number.

For the final claim, begin instead withThe expected number $T$ of triangles isso with probability tending to one $T<n$. Put $m=\lceil n^{0.7}\rceil$. The expected number of independent $m$-sets is bounded bybecause the positive term is $O(n^{0.7}\log n)$ whereas the negative term has order $n^{0.71}$.

Thus some such $G$ has fewer than $n$ triangles and no independent set of size $m$. Delete one vertex from every triangle; more than $n$ vertices remain and the graph is triangle-free. Take any induced subgraph on exactly $n$ vertices. Vertex deletion cannot increase the independence number, so the resulting graph satisfiesThis is the triangle-free graph with sub-power independence number construction.

The binomial random graph $G(n,p)$ has vertex set $[n]=\{1,\dots ,n\}$, and each unordered pair of distinct vertices is included as an edge independently with probability $p$.

Let $X_t$ count copies of the complete graph $K_t$. Each $t$-element vertex set forms a copy with probability $p^{\left(\genfrac{}{}{0ex}{}{t}{2}\right)}$, so linearity of expectation givesThe event $E_t$ is $\{X_t\ge 1\}$. By the first moment method,This is the clique count in a binomial random graph.

Fix $k\ge 1$ and sample $G$ from the Erdős-Rényi model with $p=\frac{1}{2}{n}^{-2/3}$. Let $X$ count copies of the complete bipartite graph $K_{3,3}$. By the expected subgraph count in the Erdős-Rényi model,

Put $m=\lceil n/k\rceil$ and let $Z$ count independent set of size $m$. If the chromatic number satisfies $\chi (G)\le k$, some colour class has at least $m$ vertices, so $Z\ge 1$. Moreover,because the positive part of the exponent is $O(n)$ whereas $p\left(\genfrac{}{}{0ex}{}{m}{2}\right)$ has order $n^{4/3}$. The first moment method therefore gives $\mathbb{P}(\chi (G)\le k)\le \mathbb{E}Z=o(1)$.

Using Markov inequality for $X$ and the union bound,for all sufficiently large $n$. Hence at least one such $G$ contains no $K_{3,3}$ and has $\chi (G)>k$.

Let $Y$ count unordered pairs of distinct $K_4$ copies that have a common vertex. Classifying such a pair by whether it shares one, two, or three vertices givesIndeed, after substituting $p=n^{-2/3}\log n$, these terms are respectively

The first moment method gives $\mathbb{P}(Y>0)\le \mathbb{E}Y\to 0$. Part (i) says that $X\ge 100$ with probability tending to one, so the union bound shows that, with probability tending to one, both $X\ge 100$ and $Y=0$. On that event any one hundred of the $K_4$ copies are pairwise vertex-disjoint, proving vertex-disjoint sparse clique copies in this case.